Chapter 9 - Digital PCM

Question 1

WTAD explain how an analogue voice signal is changed to a PCM signal. What is the significance of the Nyquist Sampling Rate? What are the two disadvantages of PCM?

Answer 1

How an Analogue Voice Signal is Changed to a PCM Signal

By regularly sampling the baseband signal, mapping each sample to the nearest quantisation level, and then converting these quantised levels to a binary format.

Significance of the Nyquist Sampling Rate

Because it states that the sampling frequency must be at least twice the highest frequency present in the baseband to accurately reproduce the signal and prevent aliasing .

Two Disadvantages of PCM

1. High Bandwidth Requirement: PCM requires a higher bandwidth compared to other digital modulation techniques .
2. Quantisation Noise: The difference between the continuous baseband signal and the discrete PCM output introduces quantisation noise .

Question 2

WTAD explain how Differential Pulse Code Modulation is used to digitize an analogue voice signal. What is the main problem of receiving DM or DPCM? What major benefit does DPCM have over PCM?

Answer 2

1. How DPCM is used to digitize an analogue voice signal:
   
   • DPCM transmits the difference between each sample and the preceding sample, reducing bandwidth while maintaining signal quality.
2. Main problem of receiving DM or DPCM:
   
   • Errors in transmission propagate throughout the recovered baseband signal, requiring periodic full sample resets to correct.
3. Major benefit of DPCM over PCM:
   
   • DPCM reduces the bandwidth required for transmission without increasing quantisation noise.

Question 3

A toll quality system employs 512 quantisation levels and samples 0-3 KHz voice signals at a rate of 8000 times per second. What will be the output bit rate of this A-D converter? (72 kbits/sec)

Answer 3

Frequency Range = 0-3 kHz
Sampling Rate = 8000 samples per second
Quantisation Levels = 512
No. Bits required = Log2(512) = 9 bits

Bit rate = sampling rate * no. bits
8000 * 9 = 72,000

= 72 kbps

Question 4

256 levels are used to quantise a 4KHz Signal. Assuming the sampling is at exactly the Nyquist rate what is the bit rate? (64000bps)

Answer 4

Frequency = 4 kHz
Sampling rate = 2 x 4 kHz = 8 kHz
Quantisation level = 256
bits required = log2(256) = 8 bits

bit rate = sampling rate * number of bits per sample

8 kHz * 8 bits = 64 kbps